Question

我正在使用SpringMVC开发一个REST服务，我在类和方法级别有@RequestMapping。

此应用程序当前配置为返回在web.xml中配置的错误页面jsp。

<error-page>
    <error-code>404</error-code>
    <location>/resourceNotFound</location>
</error-page>

但我希望返回自定义JSON而不是此错误页面。

我能够处理异常并返回json以获取其他异常，通过在控制器中编写它，但不确定当url根本不存在时，如何以及在何处编写逻辑以返回JSON。

    @ExceptionHandler(TypeMismatchException.class)
        @ResponseStatus(value=HttpStatus.NOT_FOUND)
        @ResponseBody
        public ResponseEntity<String> handleTypeMismatchException(HttpServletRequest req, TypeMismatchException ex) {

            HttpHeaders headers = new HttpHeaders();
            headers.add("Content-Type", "application/json; charset=utf-8");
            Locale locale = LocaleContextHolder.getLocale();
            String errorMessage = messageSource.getMessage("error.patient.bad.request", null, locale);

            errorMessage += ex.getValue();
            String errorURL = req.getRequestURL().toString();

            ErrorInfo errorInfo = new ErrorInfo(errorURL, errorMessage);
            return new ResponseEntity<String>(errorInfo.toJson(), headers, HttpStatus.BAD_REQUEST);

        }

我试过@ControllerAdvice，它适用于其他异常情况，但不适用于映射不可用的情况

@ControllerAdvice
public class RestExceptionProcessor {

    @Autowired
    private MessageSource messageSource;

    @ExceptionHandler(HttpRequestMethodNotSupportedException.class)
    @ResponseStatus(value=HttpStatus.NOT_FOUND)
    @ResponseBody
    public ResponseEntity<String> requestMethodNotSupported(HttpServletRequest req, HttpRequestMethodNotSupportedException ex) {
        Locale locale = LocaleContextHolder.getLocale();
        String errorMessage = messageSource.getMessage("error.patient.bad.id", null, locale);

        String errorURL = req.getRequestURL().toString();

        ErrorInfo errorInfo = new ErrorInfo(errorURL, errorMessage);
        return new ResponseEntity<String>(errorInfo.toJson(), HttpStatus.BAD_REQUEST);
    }

    @ExceptionHandler(NoSuchRequestHandlingMethodException.class)
    @ResponseStatus(value=HttpStatus.NOT_FOUND)
    @ResponseBody
    public ResponseEntity<String> requestHandlingMethodNotSupported(HttpServletRequest req, NoSuchRequestHandlingMethodException ex) {
        Locale locale = LocaleContextHolder.getLocale();
        String errorMessage = messageSource.getMessage("error.patient.bad.id", null, locale);

        String errorURL = req.getRequestURL().toString();

        ErrorInfo errorInfo = new ErrorInfo(errorURL, errorMessage);
        return new ResponseEntity<String>(errorInfo.toJson(), HttpStatus.BAD_REQUEST);
    }


}

Answer 1

在SpringFramework中挖掘DispatcherServlet和HttpServletBean.init（）之后，我发现它在Spring 4中是可行的。

<强> org.springframework.web.servlet.DispatcherServlet

/** Throw a NoHandlerFoundException if no Handler was found to process this request? **/
private boolean throwExceptionIfNoHandlerFound = false;

protected void noHandlerFound(HttpServletRequest request, HttpServletResponse response) throws Exception {
    if (pageNotFoundLogger.isWarnEnabled()) {
        String requestUri = urlPathHelper.getRequestUri(request);
        pageNotFoundLogger.warn("No mapping found for HTTP request with URI [" + requestUri +
                "] in DispatcherServlet with name '" + getServletName() + "'");
    }
    if(throwExceptionIfNoHandlerFound) {
        ServletServerHttpRequest req = new ServletServerHttpRequest(request);
        throw new NoHandlerFoundException(req.getMethod().name(),
                req.getServletRequest().getRequestURI(),req.getHeaders());
    } else {
        response.sendError(HttpServletResponse.SC_NOT_FOUND);
    }
}

throwExceptionIfNoHandlerFound 默认为false，我们应该在 web.xml

中启用它

<servlet>
    <servlet-name>appServlet</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>throwExceptionIfNoHandlerFound</param-name>
            <param-value>true</param-value>
        </init-param>
    <load-on-startup>1</load-on-startup>
    <async-supported>true</async-supported>
</servlet>

然后您可以使用此方法在使用 @ControllerAdvice 注释的类中捕获它。

@ExceptionHandler(NoHandlerFoundException.class)
@ResponseStatus(value=HttpStatus.NOT_FOUND)
@ResponseBody
public ResponseEntity<String> requestHandlingNoHandlerFound(HttpServletRequest req, NoHandlerFoundException ex) {
    Locale locale = LocaleContextHolder.getLocale();
    String errorMessage = messageSource.getMessage("error.bad.url", null, locale);

    String errorURL = req.getRequestURL().toString();

    ErrorInfo errorInfo = new ErrorInfo(errorURL, errorMessage);
    return new ResponseEntity<String>(errorInfo.toJson(), HttpStatus.BAD_REQUEST);
}

这允许我为没有映射的坏URL返回JSON响应，而不是重定向到JSP页面：）

{"message":"URL does not exist","url":"http://localhost:8080/service/patientssd"}

Answer 2

如果您使用的是Spring Boot，请设置以下两个属性：

private static bool IsValidPath(string input)
{
   bool isValid = false;
   try
   {
      var fullPath = Path.GetFullPath(input);
      return input.Equals(fullPath, StringComparison.InvariantCultureIgnoreCase);
   }
   catch (Exception)
   {
      isValid = false;
   }
   return isValid;
}

现在你的@ControllerAdvice注释类可以处理“NoHandlerFoundException”，如下所示。

spring.resources.add-mappings=false
spring.mvc.throw-exception-if-no-handler-found=true

请注意，仅指定此属性是不够的：

@ControllerAdvice
@RequestMapping(produces = "application/json")
@ResponseBody
public class RestControllerAdvice {

    @ExceptionHandler(NoHandlerFoundException.class)
    public ResponseEntity<Map<String, Object>> unhandledPath(final NoHandlerFoundException e) {
        Map<String, Object> errorInfo = new LinkedHashMap<>();
        errorInfo.put("timestamp", new Date());
        errorInfo.put("httpCode", HttpStatus.NOT_FOUND.value());
        errorInfo.put("httpStatus", HttpStatus.NOT_FOUND.getReasonPhrase());
        errorInfo.put("errorMessage", e.getMessage());
        return new ResponseEntity<Map<String, Object>>(errorInfo, HttpStatus.NOT_FOUND);
    }

}

，默认情况下，Spring会将未知的网址映射到/ **，因此确实没有“找不到处理程序”。

要禁用未知网址映射到/ **，您需要

spring.mvc.throw-exception-if-no-handler-found=true

这就是两个属性一起产生所需行为的原因。

Answer 3

如果您使用的是Spring 3.2或更高版本，则可以使用控制器建议（@ControllerAdvice）来处理映射错误（404）。你可以找到documentation here。看看第17.11节。例如，您可以使用此方法提供更详细的日志记录，说明您的请求绑定未与特定网址匹配的原因，或仅返回比通用404更具体的响应。

Answer 4

你可以在下面的位置返回json， / handle / 404 。

<error-page>
    <error-code>404</error-code>
    <location>/handle/404</location>
</error-page>

在web.xml中配置后，404错误将重定向到 / handle / 404 ，您可以使用此映射创建控制器并返回json结果。例如。

@RestController
@RequestMapping(value = "handle")
public class HttpErrorController {
   @RequestMapping(value = "404")
   public String handle404() {
      return "404 error";
   }
}

Spring MVC REST通过返回JSON来处理Bad Url（404）

4 个答案: